Stoichiometry and limiting reactants?

Question

1) What masses of Iron (III) Oxide and aluminum must be used to produce 15 g of iron? What is the maximum mass of aluminum oxide that could be produced?

Fe2O3 + 2AL ---> 2Fe + Al2O3

2) When aluminum burns in bromine, aluminum bromide is produced. In a certain experiment, 6.0 g of aluminum was reacted with an excess of bromine to yield 50.3g aluminum bromide. Calculate the theoretical yield and percentyield for this experiment.

Need explanation please!

Answer 1

Find the molecular masses of each compound in the reaction

Fe2O3 159.69
Al 26.98
Fe 55.85
Al2O3 101.96

15g of iron represents 15/55.85 moles of iron = 0.269 mole Fe

Look at the balanced reaction: one mole Fe2O3 produces 2 moles of Fe, so Fe2O3 needed is half the moles of Fe, or 0.134 moles of Fe2O3. 0.134 moles of Fe2O3 has a mass of 0.134*159.69 = 21.4 grams Fe2O3

The moles of Al2O3 produced are also half the moles of Fe, or 0.134 moles. The mass of Al2O3 is then 0.134*101.96 = 13.66 grams of Al2O3

The molecular mass of AlBr3 is 266.69

The reaction is Al + 3Br ---> AlBr3

6 grams of Al is 6/26.98 = 0.222 moles of Al, this will produce 0.222 moles of AlBr3. 50.3 g of AlBr3 is is 50.3/266.69 or only 0.189 moles. The reaction produces less than the maximum amount by the ratio 0.189/0.222 = 0.85. Therefore the percent yield is 85%